FrictionFriction Problem Situations

Friction

• Friction Ff is a force that resists motion•Friction involves objects in contact

with each other.•Friction must be overcome before

motion occurs.• Friction is caused by the uneven

surfaces of the touching objects. As surfaces are pressed together, they tend to interlock and offer resistance to being moved over each other.

Friction

• Frictional forces are always in the direction that is opposite to the direction of motion or to the net force that produces the motion.

• Friction acts parallel to the surfaces in contact.

Types of Friction• Static friction: maximum frictional force

between stationary objects.• Until some maximum value is reached and

motion occurs, the frictional force is whatever force is necessary to prevent motion.

• Static friction will oppose a force until such time as the object “breaks away” from the surface with which it is in contact.

• The force that is opposed is that component of an applied force that is parallel to the surface of contact.

Types of Friction• The magnitude of the static friction force Ffs

has a maximum value which is given by:

• where μs is the coefficient of static friction and FN is the magnitude of the normal force on the body from the surface.

fs s NF F

Types of Friction• Sliding or kinetic friction: frictional force

between objects that are sliding with respect to one another.• Once enough force has been applied to the

object to overcome static friction and get the object to move, the friction changes to sliding (or kinetic) friction.

• Sliding (kinetic) friction is less than static friction.• If the component of the applied force on the

object (parallel to the surface) exceeds Ffs then the magnitude of the opposing force decreases rapidly to a value Fk given by:

where μk is the coefficient of kinetic friction.

k k NF F

Types of Friction• Here you can see

that the applied force is resisted by the static frictional force Ffs (fs in the figure) until “breakaway”.

• Then the sliding (kinetic) frictional force Fk is approximately constant.

Types of Friction

• Static and sliding friction are dependent on:• The nature of the surfaces in contact.

Rough surfaces tend to produce more friction.

• The normal force (Fn) pressing the surfaces together; the greater Fn is, the more friction there is.

Types of Friction

Types of Friction

• Rolling friction: involves one object rolling over a surface or another object.

• Fluid friction: involves the movement of a fluid over an object (air resistance or drag in water) or the addition of a lubricant (oil, grease, etc.) to change sliding or rolling friction to fluid friction.

Coefficient of Friction

• Coefficient of friction (): ratio of the frictional force to the normal force pressing the surfaces together. has no units.

• Static:

• Sliding (kinetic):

n

fss F

Fμ

n

fkk F

Fμ

•The maximum frictional force is 50 N. As the applied force increases from 0 N to 50 N, the frictional force also increases from 0 N to 50 N and will be equal to the applied force as it increases.

•As the applied force increases beyond 50 N, the frictional force remains at 50 N and the 100 N block will accelerate.

•Once the static frictional force of 50 N has been overcome, only a 40 N force is needed to overcome the 40 N kinetic frictional force and produce constant velocity (a = 0 m/s2).

Horizontal Surface – Constant Speed

•Constant speed: a = O m/s2.•The normal force pressing the surfaces together is the weight; Fn = Fw

fx

fx

fx

x

FF

N0FF

amFF

amFΣ

wkfx

wkf

w

f

n

fk

FμFF

FμF

FF

FF

μ

Horizontal Surface: a > O m/s2

wkf

w

f

n

fk

wn

fx

x

fx

FμF

FF

FF

μ

FF

amFF

amFΣ

FF

Horizontal Surface: a > O m/s2

• If solving for:• Fx:

• Ff:

• a:

gmμamF

FμamF

FamF

kx

wkx

fx

amFF xf

mFF

a fx

Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)

• The frictional force is responsible for the negative acceleration.

• Generally, there is no Fx.

wkf

w

f

n

fk

wn

f

FμF

FF

FF

μ

FF

amF

Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)

• Most common use involves finding acceleration with a velocity equation and finding k:

• Acceleration will be negative because the speed is decreasing.

)ta(vv

)ta5.0()tv(xΔ

)xΔa2(vv

if

2i

2i

2f

Horizontal Surface: Skidding to a Stop or Slowing Down (a < O m/s2)

• The negative sign for acceleration a is dropped because k is a ratio of forces that does not depend on direction.

• Maximum stopping distance occurs when the tire is rotating. When this happens, a = -s·g.

• Otherwise, use a = -k·g to find the acceleration, then use a velocity equation to find distance, time, or speed.

ga

gmam

FF

FF

μw

f

n

fk

Down an Inclined Plane

Down an Inclined Plane

• Resolve Fw into Fx and Fy.

• The angle of the incline is always equal to the angle between Fw and Fy.

• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.

θsinFF

θcosFF

FF

θsinF

Fθcos

wx

wy

w

x

w

y

Down an Inclined Plane

• The force pressing the surfaces together is NOT Fw, but Fy; Fn = Fy.

or

ykf

y

f

n

fk

fx

fx

FμF

FF

FF

μ

mFF

a

amFF

amΣF

cos

)sin(

)cos()sin(

cossin

g

ag

gga

outcancelsmmass

gmgmma

Down an Inclined Plane

• For constant speed (a = 0 m/s2):

θtanμ

θtanθcosθsin

θcosFθsinF

FF

FF

μ

FF

N0FFs

m0mFF

k

w

w

y

x

n

fk

fx

fx

2fx

Down an Inclined Plane

• To determine the angle of the incline:• If moving:

• If at rest:

k1 μtanθ

s1 μtanθ

Non-parallel Applied Force on Ramp

mgmg cos

mg sin

fk

N

Suppose the applied force acts on the box, at an angle above the horizontal, rather than parallel to the ramp. We must resolve FA into parallel and perpendicular components using the angle + (FA cos ( + θ) and FA sin ( + θ)).

FA serves to increase acceleration directly and indirectly: directly by FA cos ( + θ) pulling the box down the ramp, and indirectly by FA sin ( + θ) lightening the normal support force with the ramp (thereby reducing friction).

FA

FA cos( + )

FA sin( + )

continued on next slide

Non-parallel Applied Force on Ramp

mgm

g cos

mg sin

fk

N

FA

FA cos( + )

FA sin ( + )

Because of the perp. comp. of FA, N < mg cos. Assuming FA sin( + ) is not big enough to lift the box off the ramp, there is no acceleration in the perpendicular direction. So, FA sin( + ) + N = mg cos. Remember, N is what a scale would read if placed under the box, and a scale reads less if a force lifts up on the box. So, N = mg cos - FA sin( + ), which means fk = k N = k [mg cos - FA sin( + )].

continued on next slide

Non-parallel Applied Force on Ramp

mg

mg cos

mg sin

fk

N

FA

FA cos( + )

FA sin( + )

If the combined force of FA cos( + ) + mg sin is is enough to move the box:

FA cos( + ) + mg sin

- k [mg cos - FA sin( + )] = ma

Up an Inclined Plane

Up an Inclined Plane• Resolve Fw into Fx and Fy.

• The angle of the incline is always equal to the angle between Fw and Fy.

• Fw is always the hypotenuse of the right triangle formed by Fw, Fx, and Fy.

θsinFF

θcosFF

FF

θsinF

Fθcos

wx

wy

w

x

w

y

Up an Inclined Plane

• Fa is the force that must be applied in the direction of motion.

• Fa must overcome both friction and the x-component of the weight.

• The force pressing the surfaces together is Fy.

Up an Inclined Plane

ykf

y

f

n

fk

xfa

xfa

x

yn

FμF

FF

FF

μ

mFFF

a

amFFF

amFΣ

FF

•For constant speed, a = 0 m/s2.

Fa = Fx + Ff

•For a > 0 m/s2.Fa = Fx + Ff +

(m·a)

Pulling an Object on a Flat Surface

Pulling an Object on a Flat Surface

•The pulling force F is resolved into Fx and Fy.

θsinFF

θcosFFF

Fθsin

FF

θcos

y

x

y

x

Pulling an Object on a Flat Surface

•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw minus the upward force Fy from the pulling force.•For constant speed, a = 0 m/s2.

mFF

a

amFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fx

fx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Simultaneous Pulling and Pushing an Object on a Flat Surface

Simultaneous Pulling and Pushing an Object on a Flat Surface

θsinFF

θcosFFF

Fθsin

FF

θcos

y

x

y

x

m

FFFa

amFFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fpushx

fpushx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Pushing an Object on a Flat Surface

Pushing an Object on a Flat Surface

•The pushing force F is resolved into Fx and Fy.

sinθFF

cosθFFF

Fsinθ

FF

cosθ

y

x

y

x

Pushing an Object on a Flat Surface

•Fn is the force that the ground exerts upward on the mass. Fn equals the downward weight Fw plus the upward force Fy from the pushing force.•For constant speed, a = 0 m/s2.

mFF

a

amFF

amFΣ

)FF(μF

FFF

FF

μ

FFF

N0FFF

N0FΣ

fx

fx

x

ywkf

yw

f

n

fk

ywn

wyn

y

Pulling and Tension

• The acceleration a of both masses is the same.

Pulling and Tension

• For each mass:

• Isolate each mass and examine the forces acting on that mass.

n2kf2

n1kf1

w2n2

w1n1

FμF

FμF

FF

FF

Pulling and Tension

•m1 = mass

•T1 may not be a tension, but could be an applied force (Fa) that causes motion.

amFTT

amΣF

11f21

1

Pulling and Tension

•m2 = mass

amFT

amΣF

22f2

2

Pulling and Tension

• This problem can often be solved as a system of equations:

• See the Solving Simultaneous Equations notes for instructions on how to solve this problem using a TI or Casio calculator.

amFT

amFTT

22f2

11f21

Revisiting Tension and Friction

Revisiting Tension and Friction

•For the hanging mass, m2:

•The acceleration a of both masses is the same.

•For the mass on the table, m1:

amTgm

amTF

amΣF

gmF

22

22w

2

22w

amT-F

FμF

FF

amΣF

1f

1nf

1w1n

Revisiting Tension and Friction

12

f2

12f2

2f12

mmFgm

a

amamFgm

amFamgm

Normal Force Not Associated with Weight.

• A normal force can exist that is totally unrelated to the weight of an object.

applied forcefriction

weightnormal

FN = applied force

Friction is always parallel to surfaces….

•In this case, for the block to remain in position against the wall without moving:

• the upward frictional force Ff has to be equal and opposite to the downward weight Fw.•The rightward applied force F has to be equal ad opposite to the leftward normal force FN.

F

FW

Ff

FN

(0.20)